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$$\int x\,dx^2$$

I'm having trouble comprehending this question. I can perform substitution, but when I do I come out wrong. Here's how I've happened to handle this:

$\int x\,dx^2$

$u=x^2$

$du=2x\,dx$

$dx=\frac {du}{2x}$

$\frac12 \int x\frac{du}{2x}$

$\frac 12\int\frac12\,du$

$\frac u4$

$\frac {x^2}{4}$

It says that it's wrong. Me showing my work was just a desperate attempt at solving it. I'd really like a guide through the problem please.

Also, apparently the answer is $\frac {2x^3}{3}$

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  • $\begingroup$ You substituted $\mathrm{d}x$ for $\mathrm{d}(x^2)$, but these are not equal. $\endgroup$ – nathan.j.mcdougall May 19 '15 at 1:14
  • $\begingroup$ So would that apply to my $du$? Thanks for the help by the way! $\endgroup$ – PhilTheLawyer May 19 '15 at 1:15
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    $\begingroup$ My Russian high school teacher always said "Back in Russia it was minus $50$ points for forgetting the $+C$." For us, it was minus $10$ points the first time, but minus all the points any subsequent times. I learned after the first time. In short, the answer should have a $+C$. $\endgroup$ – Peter Woolfitt May 19 '15 at 1:17
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    $\begingroup$ Indeed. These [1] [2] comics should make you laugh and remember to be in the habit of keeping those +C's (or +monkeys etc...) to avoid losing points. $\endgroup$ – JMoravitz May 19 '15 at 1:19
  • $\begingroup$ @PeterWoolfitt: glad that such teachers don't live here in Russia :) Forgetting $+C$ seems to me like a typo, if the concept of "$+C$" is understood :) $\endgroup$ – Alexey Burdin May 19 '15 at 1:21
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If $u = x^2$, then modulo a sign, $x = \sqrt{u}$. Hence the integral is equal to

$$\int \sqrt{u} \ du = \frac{2}{3}u^{3/2} \ (+ C) = \frac{2}{3}x^3 \ (+ C)$$

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An alternative approach: $d(x^2) = 2x\,dx$ so you have

$$\int x\,d(x^2) = \int 2x^2 \,dx = \frac{2}{3}x^3+C.$$

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Everything was going fine until $$\mathrm{d}x=\frac{\mathrm{d}u}{2x}$$. Instead, letting $u=x^2\implies x=\sqrt{u}$ $$\int\!x\,\mathrm{d}(x^2)=\int\!\sqrt{u}\,\mathrm{d}u$$ $$=\frac{2}{3}\sqrt{u}^3+c$$ $$=\frac{2}{3}x^3+c$$

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  • $\begingroup$ I see what I did! Thank you so much!! $\endgroup$ – PhilTheLawyer May 19 '15 at 1:18
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you can solve this integration by another method. As we are familiar with integrating with dx instead of $dx^2$, so it will be easier if we translate it to an integral with dx.

Let $y = \int x\,dx^2$ so $\frac{dy}{dx^2}=x$ then we divide by $dx$ to get $\frac{\frac{dy}{dx}}{\frac{dx^2}{dx}} = x$. $\frac{dx^2}{dx}=2x$, that mean $\frac{y'}{2x}=x$ so $\frac{dy}{dx}=2x^2$ so $y = \frac{2x^3}{3}+k$

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It is very simple:

$$\int xdx^2 = \int \sqrt{x^2}dx^2=\frac{2}{3}x^3 + c$$

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